A277858 - OEIS

%I #31 Nov 01 2019 20:49:30

%S 1,2,4,8,16,10,9,18,17,24,20,22,44,40,36,30,27,25,23,21,19,28,26,32,

%T 29,31,34,37,74,67,60,54,49,45,41,42,38,35,70,63,57,50,55,110,99,90,

%U 81,73,66,72,65,59,64,58,53,48,52,47,43,39,78,71,142,100,101,91,82,80,88,96,87,79,86,92,83,75,68,62,56,51,102

%N The absolute difference between a(n) and a(n+1) is visible in a(n).

%C "Is visible in" means "is a substring of". (For example, if a(n) = 123, the absolute difference may be any of { 1, 2, 3, 12, 23, 123 } but not 13.)

%C The sequence was not extended with the smallest unused integer so far that does not lead to a contradiction, as this method would produce the finite sequence: 1, 2, 4, 8, 16, 10, 9, 18, 17, 24, 20, 22, 44, 40, 36, 30, 27, 25, 23, 21, 19, 28, 26, 32, 29, 31, 34, 37, 74, 67, 60, 54, 49, 45, 41, 42, 38, 35, 70, 63, 57, 50, 55, 110, 99, 90, 81, 73, 66, 72, 65, 59, 64, 58, 53, 48, 52, 47, 43, 39, 78, 71, 142, 100, 101, 91, 82, 80, 88, 96, 87, 79, 86, 92, 83, 75, 68, 62, 56, 51 and 46. This 46 has no successor as 40, 42 and 92 are already in the sequence.

%C Instead, a "pseudo-backtracking" method was applied: extend the sequence with the smallest unused integer so far that does not lead to a contradiction iff the said integer has itself a successor not yet in the sequence. With this method the sequence becomes infinite and doesn't stop anymore with 46 [46 was replaced by 102 (which is 51, the previous term, times 2)].

%C It looks like the sequence is not a permutation of the integers > 0. Indeed, after computing 40000 terms, the set of the unused integers so far hasn't evolved since the first 20000 terms of the sequence were computed: 3, 5, 6, 7, 11, 12, 13, 14, 15, 33, 46, 61. Question: could this 12-integer set be reduced (to 11, 10 or fewer elements) starting the sequence with another a(1) than 1?

%C For the starting value a(1) = 3 the set of unused numbers becomes { 1, 2, 4, 5, 7, 8, 11, 71 } (8 elements), for a(1) = 5 it becomes { 1, 2, 3, 4, 6, 7, 8, 33, 55 } (9 elements). - _M. F. Hasler_, Nov 04 2016

%C It is easy to prove that this is not a permutation of the positive integers: indeed, for example the number 5 can never occur, since it has no possible successor (10 being already used). There are several variants of this sequence, some of which are permutations: The condition "is a substring of" could be relaxed to (strongest to weakest) "digits of |a(n+1)-a(n)| are a subsequence of the digits of a(n)" (e.g., a step of 13 would be allowed after 123), "... are a sub-multiset of ..." (a step of 21 would be allowed after 123), or "...are a subset of ..." (a step of 11 would be allowed after 123). The last variant (which is a permutation) is A276070. - _M. F. Hasler_, Nov 05 2016

%H Jean-Marc Falcoz, <a href="/A277858/b277858.txt">Table of n, a(n) for n = 1..10000</a>

%e The absolute difference between 1 and 2 is 1, which is visible in 1; the abs diff between 2 and 4 is 2 (visible in "2"); the abs diff between 4 and 8 is 4 (visible in "4"); the abs diff between 8 and 16 is 8 (visible in "8"); the abs diff between 16 and 10 is 6 (visible in the "6" of "16"); the abs diff between 10 and 9 is 1 (visible in the "1" of "10"), etc.

%o (PARI) u=[a=1];(S(a,d=[a])=while(d[1]>9,for(k=10,d[1],d=concat(d,d[1]%k);k=k*10-1);d=concat(d[1..1]\10,d));Set(select(t->t>0&&!setsearch(u,t),concat(apply(x->[a-x,a+x],Set(d))))));for(n=1,99,print1(a",");u=setunion(u,[a]); for(k=1,#a=S(a),S(a[k])&&(a=a[k])&&break)) \\ _M. F. Hasler_, Nov 04 2016

%Y Cf. A276070.

%K nonn,base

%O 1,2

%A _Eric Angelini_ and _Jean-Marc Falcoz_, Nov 02 2016